Q. 3.1 - A First Course in Probability

Question

Show that if $P\left(A\right)>0$, then

$P(AB\mid A)\ge P(AB\mid A\cup B)$

Step-by-Step Solution

Answer

We proved that $P(AB\mid A)\ge P(AB\mid A\cup B)$ by applying conditional probability as $P\left(A\right)>0$.

Step 1: Given Information

If $P\left(A\right)>0$, We have to prove that $P(AB\mid A)\ge P(AB\mid A\cup B)$.

Step 2: Explanation

$AB\subseteq (A\cup B)\Rightarrow AB\cap (A\cup B)=AB$

The conditional probability $P[AB\mid (A\cup B\left)\right]$ can be reduced to

$P[AB\mid (A\cup B\left)\right]=\frac{P[AB\cap (A\cup B\left)\right]}{P(A\cup B)}=\frac{P\left(AB\right)}{P(A\cup B)}$

Again, from set arithmetic

$A\subseteq A\cup B \Rightarrow P\left(A\right)\le P(A\cup B)$

Finally

$P(AB\mid A)=\frac{P\left(AB\right)}{P\left(A\right)}\ge \frac{P\left(AB\right)}{P(A\cup B)}=P[AB\mid (A\cup B\left)\right]$

Step 3: Final Answer

$P(AB\mid A)$ - is the percentage of $A$ that is in $A B$

$P[AB\mid (A\cup B\left)\right]$ - is the percentage of $A\cup B$ that is in $A B$.

 And since $A\cup B$ is larger, $P[AB\mid (A\cup B\left)\right]\le P(AB\mid A)$.